The problem involves solving the equation $\cos x = x^3$ and proceeds as follows:

(a) Prove that the equation has at least one real root:

1. Define the function $f(x) = \cos x - x^3$.
2. $f(x)$ is continuous on the interval $[0, 1]$.
3. Evaluate $f(0) = \cos(0) - 0^3 = 1$.
4. Evaluate $f(1) = \cos(1) - 1^3 \approx -0.46$.
5. Since $f(0) > 0$ and $f(1) < 0$, by the Intermediate Value Theorem, there exists a root $c \in (0, 1)$ where $f(c) = 0$, or $\cos x = x^3$ has a root.

(b) Use your calculator to find an interval of length 0.01 that contains a root:

Your answer of $(0.845, 0.855)$ is likely an attempt at this, but it is marked incorrect. You might need to recalculate this using a more accurate method or interval of approximation, perhaps refining the bounds using a tool like Newton's Method or the Bisection Method, or using the calculator’s root-finding function.

Would you like help with recalculating the interval or any other details?

Here are 5 follow-up questions based on this:

1. What is the importance of the Intermediate Value Theorem in this problem?
2. How would you apply Newton's Method to refine the root's approximation?
3. Can you find the derivative of $f(x) = \cos x - x^3$ to help with Newton's Method?
4. Why does $f(x)$ need to be continuous to apply the Intermediate Value Theorem?
5. How can the Bisection Method be used to narrow down the interval further?

Tip: When using a calculator to find roots, ensure it is set to the correct mode (radians for trigonometric functions) to avoid incorrect results.